# Tag Info

## Hot answers tagged electric-circuits

11

$\def\vE{{\vec{E}}}$ $\def\vD{{\vec{D}}}$ $\def\vB{{\vec{B}}}$ $\def\vJ{{\vec{J}}}$ $\def\vr{{\vec{r}}}$ $\def\vA{{\vec{A}}}$ $\def\vH{{\vec{H}}}$ $\def\ddt{\frac{d}{dt}}$ $\def\rot{\operatorname{rot}}$ $\def\div{\operatorname{div}}$ $\def\grad{\operatorname{grad}}$ $\def\rmC{{\mathrm{C}}}$ $\def\rmM{{\mathrm{M}}}$ $\def\ph{{\varphi}}$ ...

5

The electric field assigns a single vector quantity to each point in space (specifically, the direction in which a positive test charge would accelerate if it popped into existence at that point, assuming it didn't perturb the setup creating the field in the first place). I believe that the difficulty of this question arises from an ambiguity in the problem ...

4

For a constant potential on the capacitor, there is no B-field and that is the case usually considered for this calculation. When charging a capacitor, the currents will generate a B-field and there is stored energy in that field (same as for an inductor). But once the charging stops, the B-field will "collapse" and cause currents to flow in the wires, ...

3

You are correct, that while charging a capacitor there will be a magnetic field present due to the change in the electric field. And of course $B$ contains energy as pointed out. However: As the capacitor charges, the magnetic field does not remain static. This results in electromagnetic waves which radiate energy away. The energy put into the magnetic field ...

3

You can reconcile both trains of thought by reconsidering your thoughts about pushing:- For DC circuits without changing magnetic fields, the voltage is the energy gained or lost per unit charge in moving from one position to another, say from the positive to the negative terminal of the battery. What a battery does, is that it creates certain ...

2

Note that the voltage induced by the changing magnetic field is directional. To reduce the resulting currents, you only need to increase resistivity in the direction the current would flow. That's what laminations do. Laminations are thin sheets of metal that conduct electricity (as a unintentional side effect of having desirable magnetic properties). ...

2

Am I correct that you can rephrase your question to 'electrons move so slow, how come that when I flip the light switch the light comes on basically instantly?'? It's true that the electrons travel very slowly. But these electrons don't have to travel across the wire to power your light bulb. In electromagnetism, we have the continuity equation $\nabla J = ... 2 The internal resistance is 0. An ideal voltage source is there in order to supply current at a constant voltage. The amount of current flowing through it is entirely determined by the outside circuit. Imagine a short circuit with an ideal voltage source. The current would skyrocket to infinity (Obviously any real voltage source would soon run out of ... 2 If you just plug in your suggested solution, you get $$\frac d{dt} A\cos(\omega t + \phi)+\frac 1{\tau}A\cos(\omega t + \phi)=\frac{V_{in}}\tau\sin(\omega t)\\ -A\omega \sin(\omega t + \phi)+\frac 1{\tau}A\cos(\omega t + \phi)=\frac{V_{in}}\tau\sin(\omega t)$$ Now you should be able to use the function sum formulas to solve for$\phi$and$\frac A{V_{in}} ...

2

A general strategy for these questions is to start at the battery and trace the current through the circuit. So, starting from the batter, we can see that the entire current passes through $R_1$. After that, the current hits a split (at the top of the circuit in your drawing), where part of it goes to the left through $R_2$ and the other part of it goes to ...

1

I think you have some misconceptions about voltage. You mention "net voltage" but voltage is always a difference in electric potential. In a circuit, that means you never talk about a "net voltage" or a voltage at a certain point. Voltage is always meant to be read as "the voltage between two points" or "the voltage at A with respect to B." It is never ...

1

You can't talk about "correct" without a definition of what you are trying to achieve. However, you can rule out some obviously bad arrangements if the placement of the meters keeps the circuit from working, even though we have to guess at what the intended "working" is. This is a poorly specified problem, or there is context surrounding it that you ...

1

Would there be any current flowing though an ideal voltage source? Yes. An ideal voltage source will produce whatever current is needed to maintain its voltage. This current will flow in a complete circuit through the source and through whatever network is connected to the source. I am asking this more specifically when you are analyzing a circuit ...

1

Initially when you attach the capacitor to the battery, said battery will act to create an electric field within the wire. On the side of the negative terminal this field will point perpendicular to the cross section of the wire toward the terminal of the battery (electric field points toward negative charge). On the side of the positive terminal the field ...

1

$20V$ is not normally dangerous. You may not even feel it. It is not the voltage that causes danger, but rather the current it generates through your body. Anything over about $10mA$ will be unpleasant, above $50mA$ it gets dangerous. The current is determined by both the voltage and the resistance of the body. That resistance depends on how you touch the ...

1

Electromagnetic force is not propagated by electrons, it is propagated by photons. By definition these travel at the speed of light (in the material). Impedance and capacitance play a part in how quickly the system responds to you turning it on / connection a battery, but are generally very small in a plain wire. The electrons are moved by electromagnetism ...

1

I don't understand the question and the diagram associated with it, partly because it contradicts with something I learned. I was told that if I have a DC current in a straight current, by Ohm's law $\mathbf{J} = \sigma \mathbf{E}$, the electric field lines must be all directed along the current. So I don't see a way by which surface charge can build up on ...

1

Here's a simple explanation: the wire is assumed initially to have a net charge of zero. That is, for every electron, there is a proton. Because charge is conserved, and you haven't provided any mechanism to add or remove charge from some external mechanism, there must be an equal number of positive charges (protons missing an electron, holes) and negative ...

1

Yes, until energy is conserved. Kirchhoff's law is amongst great basic laws which help in constructing electronic physics. Kirchhoff's law is based on the Law of Conservation and we also know that energy is always conserved according to our latest experiments and analysis. If in future it gets false then we must once have to think about the applicability ...

Only top voted, non community-wiki answers of a minimum length are eligible